Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {

        boolean local = false;
        if (args.length > 0 && args[0].equals("LOCAL"))
            local = true;

        new Main(local).go();
    }

    boolean LOCAL;
    PrintWriter out;
    Reader in;
    BufferedReader br;

    Main(boolean local) throws IOException {

        this.LOCAL = local;

        try {

            //br = new BufferedReader( new FileReader("input.txt") );
            in = new Reader("input.txt");
            out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
        }
        catch (Exception e) {

            //br = new BufferedReader( new InputStreamReader( System.in ) );
            in = new Reader();
            out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
        }
    }

    void go() throws IOException {

        int t = 1;

        if (this.LOCAL)
            t = in.nextInt();

        while (t > 0) {

            solve();
            t--;

            if (LOCAL)
                out.println();

        }

        out.flush();
        out.close();
    }


    int inf = 2000000000;
    int mod = 1000000007;
    double eps = 1e-9;

    int n;
    int m;

    //ArrayList<Integer>[] g = new ArrayList[300010];

    void solve() throws IOException {

        int n = in.nextInt();
        int m = in.nextInt();

        int[][] a = new int[n][m];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                a[i][j] = in.nextInt();
        }

        int cnt = 0;
        String[] ans = new String[1000000];
        int pos = 0;

        if (m >= n) {

            for (int i = 0; i < n; i++) {

                int min = 1000;
                for (int j = 0; j < m; j++)
                    min = Math.min(a[i][j], min);

                for (int j = 0; j < m; j++)
                    a[i][j] -= min;

                cnt += min;

                for (int j = 0; j < min; j++)
                    ans[pos++] = "row " + (i+1);
            }

            for (int i = 0; i < m; i++) {

                int min = 1000;
                for (int j = 0; j < n; j++)
                    min = Math.min(a[j][i], min);

                for (int j = 0; j < n; j++)
                    a[j][i] -= min;

                cnt += min;

                for (int j = 0; j < min; j++)
                    ans[pos++] = "col " + (i+1);
            }
        }
        else {

            for (int i = 0; i < m; i++) {

                int min = 1000;
                for (int j = 0; j < n; j++)
                    min = Math.min(a[j][i], min);

                for (int j = 0; j < n; j++)
                    a[j][i] -= min;

                cnt += min;

                for (int j = 0; j < min; j++)
                    ans[pos++] = "col " + (i+1);
            }

            for (int i = 0; i < n; i++) {

                int min = 1000;
                for (int j = 0; j < m; j++)
                    min = Math.min(a[i][j], min);

                for (int j = 0; j < m; j++)
                    a[i][j] -= min;

                cnt += min;

                for (int j = 0; j < min; j++)
                    ans[pos++] = "row " + (i+1);
            }

        }

        int max = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                max = Math.max(max, a[i][j]);

        if (max > 0)
            out.println(-1);
        else {
            out.println(cnt);
            for (int i = 0; i < pos; i++)
                out.println(ans[i]);
        }


    }

    class Pair implements Comparable<Pair>{

        int a;
        int b;

        Pair(int a, int b) {

            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair p) {

//            if (a > p.a) return 1;
//            if (a < p.a) return -1;
            if (b > p.b) return -1;
            if (b < p.b) return 1;
            return 0;
        }
    }

    class Item {

        int a;
        int b;
        int c;

        Item(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

    }

    class Reader {

        BufferedReader  br;
        StringTokenizer tok;

        Reader(String file) throws IOException {
            br = new BufferedReader( new FileReader(file) );
        }

        Reader() throws IOException {
            br = new BufferedReader( new InputStreamReader(System.in) );
        }

        String next() throws IOException {

            while (tok == null || !tok.hasMoreElements())
                tok = new StringTokenizer(br.readLine());
            return tok.nextToken();
        }

        int nextInt() throws NumberFormatException, IOException {
            return Integer.valueOf(next());
        }

        long nextLong() throws NumberFormatException, IOException {
            return Long.valueOf(next());
        }

        double nextDouble() throws NumberFormatException, IOException {
            return Double.valueOf(next());
        }


        String nextLine() throws IOException {
            return br.readLine();
        }

    }

}